# This expression can be made as small as we wish by

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Unformatted text preview: rantees there is an such that a solution exists and is unique for . The maximal interval of existence is the largest interval of time that includes for which a solution to [1] exists. Example. The initial value problem . is locally Lipschitz on initial condition to find and Separate variables and apply the . The maximal interval of existence is . Note that Theorem 3.17 (Maximal Interval of Existence). Let Lipschitz. Then there is a maximal open interval value problem [1] has a unique solution . Proof. Let is not in the domain of . ■ be open and containing be locally such that the initial . Theorem 3.10 guarantees there is a ball , a time interval and a unique solution of [1] such that . Let Then and we can apply theorem 3.10 again. Let Then there is a ball , a time interval and a unique solution such that . . is not empty and uniqueness implies on Continue this construction both backwards and forwards in time. Note that every endpoint lies in . The maximal interval of existence is the union of all such intervals. Let be the unique solution constructed. 17 Chapter 3 Suppose that has a closed endpoint, for example . Then and the solution could be extended further, contradicting the claim that is maximal. Therefore is open. □ is open. ■ Example (continued). Notice that Theorem 3.18 (Escape from any compact set). Suppose is an open set and is locally Lipschitz. Let be the maximal interval of existence for [1]. If is finite, then for any compact set there is a such that . Similarly, if is finite, then for any compact there is a such that . Idea of Proof. If is finite and , and extend using theorem 3.10. sketch , , , then we can define , with , extension □ , Example (continued) sketch -, solution escapes from any compact set in or as Corollary 3.19. If does not exist or is finite then either -, both branches of or as .■ The Proof. If the limit exists it cannot be in , since then the solution could be extended. However, every point for . Therefore .□ Example (continued). sketch -, - indicate or and the flow .■ Example. The initial value problem . is locally Lipschitz on . is the solution corresponding to separate variables and apply the initial condition to find Assume . For sketch interval of existence is -, and -, vertical asymptote, does not exist. ■ . The maximal ,...
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## This document was uploaded on 02/24/2014 for the course MATH 512 at Washington State University .

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